https://github.com/sulmar/dsm.sql

Przykłady ze szkolenia SQL dla DSM
https://github.com/sulmar/dsm.sql

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Przykłady ze szkolenia SQL dla DSM

• Host: GitHub
• URL: https://github.com/sulmar/dsm.sql
• Owner: sulmar
• Created: 2017-08-31T06:54:36.000Z (almost 9 years ago)
• Default Branch: master
• Last Pushed: 2022-06-13T06:40:58.000Z (about 4 years ago)
• Last Synced: 2025-09-06T22:47:29.578Z (10 months ago)
• Topics: sql
• Language: TSQL
• Size: 378 KB
• Stars: 0
• Watchers: 1
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md

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README

          
# Podstawy języka SQL 

Z użyciem przykładowej bazy danych **Sakila**

https://github.com/ivanceras/sakila/tree/master/sql-server-sakila-db

### Wybór bazy danych

~~~ sql

use sakila

~~~

### Komentarze

- Komentarz pojedycznej linii

~~~ sql

-- komentarz

~~~

- Komentarz blokowy

~~~ sql

/* 

   komentarz

   blokowy

*/

~~~

### Pobranie danych z tabeli

~~~ sql

select * from customer

~~~ 

### Wybór pól

~~~ sql

select first_name, last_name, email from customer

~~~

### Aliasy kolumn

~~~ sql

select 

	first_name as imie, 

	last_name as nazwisko, 

	email

from 

	customer

~~~

### Filtrowanie

- Filtrowanie z uzyciem operatorów logicznych

~~~ sql

select 

	first_name as imie, 

	last_name as nazwisko, 

	email,

	active

from 

	customer

where

	(first_name = 'mary' or first_name = 'john' or first_name = 'adam')

	and active = 0

~~~

- Filtrowanie na podstawie zbioru

~~~ sql

select 

	first_name as imie, 

	last_name as nazwisko, 

	email,

	active

from 

	customer

where

	first_name in ('mary', 'john', 'adam')

~~~

## Sortowanie

- sortowanie po jednej kolumnie

~~~ sql

select * from customer order by first_name

~~~

- sortowanie po kilku kolumnach

~~~ sql

select * from customer order by first_name desc, last_name desc

~~~

## Funkcje tekstowe

- operacje na tekstach

~~~ sql

select 

	first_name,

	lower(first_name),

	left(first_name, 3), 

	right(last_name, 3),

	SUBSTRING(first_name, 3, 2)

from customer

~~~

## Funkcje agregujące

- liczność rekordów

~~~ sql

select count(*) from customer where active = 1

~~~

- liczność wartości

~~~ sql

select count(first_name) as quantity from customer where active = 1

~~~

- suma 

~~~ sql

select sum(amount) as total from payment

~~~

- ilość unikalnych

~~~ sql

select count(distinct first_name) from customer

~~~

- obliczanie średniej

~~~ sql

select avg(isnull(amount,0)) from payment

~~~

## Złączenia

- złączenie wewnętrzne (inner join)

~~~ sql

select 

	customer.first_name,

	customer.last_name,

	[address].[address],

	customer.address_id,

	[address].city_id

 from customer

	inner join address 

		on customer.address_id = address.address_id  -- warunek złączenia

~~~

- złączenie wielu tabel

~~~ sql

select 

	customer.first_name,

	customer.last_name,

	[address].[address],

	city.city,

	country.country

 from customer

	inner join address 

		on customer.address_id = address.address_id  -- warunek złączenia

	inner join city

		on address.city_id = city.city_id

	inner join country

		on city.country_id = country.country_id

~~~

- Ilość klientów w poszczególnych państwach

~~~ sql

select 

	country.country,

	count(*) as [count]

 from customer

	inner join address 

		on customer.address_id = address.address_id  -- warunek złączenia

	inner join city 

		on address.city_id = city.city_id

	inner join country 

		on city.country_id = country.country_id

group by country.country

~~~

- Wyświetlenie filmów i ich kategorii

~~~ sql

select 

	film.title,

	category.name

 from film

	inner join film_category

		on film.film_id = film_category.film_id

	inner join category

		on film_category.category_id = category.category_id

~~~

## Zadania

- Z jakiej kategorii najczęściej klienci wypożyczają filmy?

~~~ sql

select 

	category.name,

	count(*) as rent_count

 from rental

	 inner join inventory 

		on rental.inventory_id = inventory.inventory_id

     inner join film_category 

		on film_category.film_id = inventory.film_id

	inner join category

		on film_category.category_id = category.category_id

group by

	category.name

~~~

## Podzapytania

- podzapytanie

~~~ sql

select sum(rent_count) from 

(select 

	category.name,

	count(*) as rent_count

 from rental

	 inner join inventory 

		on rental.inventory_id = inventory.inventory_id

     inner join film_category 

		on film_category.film_id = inventory.film_id

	inner join category

		on film_category.category_id = category.category_id

group by

	category.name) as subquery

~~~ 

### Zadanie

- Który z aktorów najczęściej występuje w filmach?

~~~ sql

select top(1)

  actor.first_name,

  actor.last_name,

  count(*) as actor_count

from film

 inner join film_actor 

	on film.film_id = film_actor.film_id

 inner join actor

	on film_actor.actor_id = actor.actor_id

group by

	actor.first_name, actor.last_name

order by count(*) desc

~~~

- Które filmy generują największe przychody (top 10)?

~~~ sql

select 

	top(10)

	film.title,

	sum(payment.amount) as total

 from payment

	inner join rental

		on payment.rental_id = rental.rental_id

	inner join inventory

		on rental.inventory_id = inventory.inventory_id

    inner join film

		on inventory.film_id = film.film_id

group by film.title

order by total desc

~~~

- W jakim kraju, której kategorii najczęściej są wypożyczane filmy?

~~~ sql

select 

   cat.name,

   ctr.country,

   count(r.rental_id) as rent_count,

   sum(p.amount) as total

from rental as r

	inner join customer as c

		on r.customer_id = c.customer_id

	inner join address as a

		on c.address_id = a.address_id

	inner join city as ct

		on a.city_id = ct.city_id

	inner join country as ctr

		on ct.country_id = ctr.country_id

	inner join inventory as i

		on r.inventory_id = i.inventory_id

	inner join film as f

		on i.film_id = f.film_id

	inner join film_category as fc

		on fc.film_id = f.film_id

	inner join category as cat

		on cat.category_id = fc.category_id

	inner join payment as p

		on p.rental_id = r.rental_id

group by

   cat.name,

   ctr.country

order by

   cat.name,

   ctr.country,

   rent_count desc

~~~

- Jaki jest średni czas wypożyczenia filmów w poszczególnych kategoriach?

~~~ sql

select 

	category.name,

	avg(datediff(hour, rental_date, return_date) / 24.0) as averange 

from rental

	inner join inventory 

		on rental.inventory_id = inventory.inventory_id

	inner join film

		on inventory.film_id = film.film_id

	inner join film_category

		on film_category.film_id = film.film_id

	inner join category

		on film_category.category_id = category.category_id

group by category.name

~~~

## Modyfikacja danych

### Wstawienie danych

- Wstawienie pojedynczego rekordu

~~~ sql

insert into customer  

	values(1, 'MARCIN', 'SULECKI', 'marcin.sulecki@sulmar.pl', 605, 1, getdate(), getdate())

~~~

- Wstawienie wielu rekordów

~~~ sql

 -- wstawienie wielu rekordów

insert into city 

	values ('Poznań', 76, getdate()), 

		   ('Warszawa', 76, getdate()), 

		   ('Niepruszewo', 76, getdate())

~~~

### Zadanie

1. dodaj nową kategorię 'Educational'

~~~ sql

insert into category

	values('Educational', getdate())

~~~

1. dodaj nowy film np. 'SQL in Action' w kategorii 'Educational'

~~~ sql

insert into film (title, language_id, rental_duration, rental_rate, length, replacement_cost, last_update)

	values ('SQL in Action', 1, 5, 4.99, 120, 20, getdate())

~~~

1. dodaj 2 kopie filmu dla sklepu Store 1 i 1 kopię filmu do sklepu Store 2

~~~ sql

insert into inventory

	values (1002, 1, GETDATE()), (1002, 1, getdate()), (1002, 2, getdate())

~~~

1. Wypozycz film 'SQL in Action' ze sklepu Store 2

~~~ sql

insert into rental 

 values(getdate(), 4583, 599, null, 1, GETDATE())

~~~

1. Zwróć wypozyczony film

~~~ sql

update rental 

	 set return_date = GETDATE()

where

	rental_id = 16051

~~~

1. które z filmów są obecnie wypożyczone?

~~~ sql

select distinct

	film.title

from rental

	inner join inventory

		on rental.inventory_id = inventory.inventory_id

	inner join film

		on inventory.film_id = film.film_id

where

	rental.return_date is null 

~~~

~~~ sql

insert into inventory

	values (1002, 1, GETDATE()), (1002, 1, getdate()), (1002, 2, getdate())

~~~

## Złączenia zewnętrzne

### Lewe złączenie zewnętrzne

~~~ sql

select 

	rental.rental_date,

	rental.customer_id,

	customer.first_name,

	customer.last_name,

	customer.customer_id

from customer

	left outer join rental

		on rental.customer_id = customer.customer_id

~~~

### Prawe złączenie zewnętrzne

~~~ sql

select 

	rental.rental_date,

	rental.customer_id,

	customer.first_name,

	customer.last_name,

	customer.customer_id

from rental

	right outer join customer

		on rental.customer_id = customer.customer_id

~~~

- Zastosowanie

~~~ sql

select 

	customer.first_name,

	customer.last_name,

	count(rental.rental_id) as rent_count

from rental

	right outer join customer

		on rental.customer_id = customer.customer_id

group by customer.first_name, customer.last_name

order by rent_count 

~~~

### Pełne złączenie zewnętrzne

~~~ sql

select 

	rental.rental_date,

	rental.customer_id,

	customer.first_name,

	customer.last_name,

	customer.customer_id

from rental

	full outer join customer

		on rental.customer_id = customer.customer_id

~~~

## Zadanie

1. w ktorych 3 miastach mamy najwiecej klientow?

~~~ sql

select 

	top(3)

    city.city

from customer

	inner join address

		on customer.address_id = address.address_id

	inner join city

		on address.city_id = city.city_id

group by city.city

order by count(*) desc

~~~

2. w jakich miastach mamy sklepy?

## Suma zbiorów

- w jakich miastach mamy przeprowadzić akcję bilboardową? Na podst. 3 miast, w których mamy najwiecej klientów i lokalizacji sklepów.  

~~~ sql

select city from

(select 

	top(3)

    city.city

from customer

	inner join address

		on customer.address_id = address.address_id

	inner join city

		on address.city_id = city.city_id

group by city.city

order by count(*) desc) as city_of_customers

union

select distinct city.city from store	

	inner join address

		on store.address_id = address.address_id

	inner join city

		on address.city_id = city.city_id

 ~~~       

## Kolejność przetwarzania zapytania

- jakie filmy w jezyku angielskim wypozyczone byly wiecej niz 3 razy?

~~~ sql

select 

	film.title,

	count(rental.rental_id) as rental_count  

from rental

	inner join inventory 

		on rental.inventory_id = inventory.inventory_id

	inner join film

		on inventory.film_id = film.film_id

where film.language_id = 1

group by film.title

having COUNT(*) > 3

order by rental_count desc

~~~

- Kolejność

1. FROM

2. WHERE

3. GROUP BY

4. HAVING

5. SELECT

6. ORDER BY

## Zadania

- jaki jest średni czas wypożyczenia filmów i dochód w podziale na kategorie?

~~~ sql

select 

	c.name,

	avg(datediff(day, rental.rental_date, rental.return_date)) as average,

	sum(p.amount) as total

from rental

	inner join inventory 

		on rental.inventory_id = inventory.inventory_id

	inner join film as f

		on f.film_id = inventory.film_id

	inner join film_category as fc

		on fc.film_id = f.film_id

	inner join category as c

		on c.category_id = fc.category_id

	inner join payment as p

		on p.rental_id = rental.rental_id

group by c.name

order by c.name		

~~~

- znajdz wszystkich klientów, którzy mają kod pocztowy zaczynający się od 30

~~~ sql

select 

	customer.first_name + ' ' + customer.last_name as fullname

 from address

	inner join customer 

		on address.address_id = customer.address_id

where address.postal_code like '30%'

~~~

## Tworzenie tabel

### Utworzenie tabeli

Utwórz tabelę finish_goods zawierającą następujące kolumny:

- finish_good_id (PK)

- code (10 znakow)

- name (do 40 znakow)

~~~ sql

create table dbo.finish_good 

(

	finish_good_id int not null primary key,

	code char(10) not null,

	name varchar(40) null

)

~~~

- Wypełnienie tabeli danymi

~~~ sql

insert into finish_good 

   values  

(1	, 'FA00357B05'	, 'Vitaminas A, D3, E y K1 premix  ' ),

(2	, 'FA00355P01'	, 'Premix Vit nutr med FT117002EU'),

(3	, 'FA00356B05'	, 'Premix Vit A, D3, E, K1 nutr m'),

(4	, 'FA00762PD1'	, 'Vit A, D, RRR-E y K premix'),

(5	, 'FA00380P01'	, 'Premix sulfato de cobre Kosher'),

(6	, '5070010RES'	, 'Citidin 5-monofosfato acido'),

(7	, '5070009RES'	, 'Adenosin 5-monofosfato acido'),

(8	, 'FA01056P01'	, 'Premix Vitam hidro y cobre 4'),

(9	, 'FA00776P01'	, 'WS Premix Vitaminas y Cobre 2'),

(10	, 'FA01040B05'	, 'Premix Vit A,D2,E,K1 Isomil In'),

(11	, 'FA00758P01'	, 'Premix Vit hidro ENG2 (16910)'),

(12	, 'FA00763P01'	, 'Premix de Vit 25913 FT117008EU'),

(13	, 'FA00923B05'	, 'OS Premix Vit China FT101775EU'),

(14	, 'FA00754B05'	, 'OS Premix de Vit 1 FT101494EU'),

(15	, 'FA01027P01'	, 'Prem de vit ES29435 FT101572EU'),

(16	, 'FA01035B05'	, 'Premix de vita OS 2 FT111223EU'),

(17	, '5070016RES'	, 'disodio Uridin 5-monofosfato'),

(18	, 'FA00333P01'	, 'Premix de vit WS 3 FT111233EU'),

(19	, 'FA00777A01'	, 'WS Premix total comfort'),

(20	, 'FA00168P01'	, 'Premix de vit HL10 FT101571EU'),

(21	, 'FA00749B05'	, 'Premix Vit/Min ENG2 FT091634EU'),

(22	, 'FA00766P01'	, 'Premix Vit/Min LF 46810'),

(23	, 'FA00765B05'	, 'Premix de vit EUOS FT117017EU'),

(24	, 'FA00367P01'	, 'WS Premix vitaminas y cobre 1'),

(25	, '5070014RES'	, 'disodio guanosin 5-monofosfato'),

(26	, 'FA00189P01'	, 'Vita Premix 22278 (FT112631EU)'),

(27	, 'FA01054B05'	, 'Premix Vitaminas liposoluble 4'),

(28	, 'FA02326P01'	, 'Premix Vitaminas hidrosolu 6'),

(29	, 'FA01842P01'	, 'Premix Vitaminas Hidrosolubl 7'),

(30	, 'FA00952B05'	, 'Premix Vitaminas Liposoluble 5'),

(31	, 'FA01841P01'	, 'Premix Vitaminas Hidrosolubl 9'),

(32	, 'FA00087B05'	, 'Premix vitaminas A,D3,E,K EEC'),

(33	, '5049006RES'	, 'Premix Hierro Sulfat FT127016B'),

(34	, 'FA04622B05'	, 'OS VITAMIN PREMIX 7')

~~~

- Wyświetl produkty rozpoczynające się na FA

~~~ sql

select distinct left(code, 7)  as code from finish_good where left(code, 7) like 'FA%'

~~~

- Wyświetl ilość produktów gotowych z grupy B05

~~~ sql

select count(*) from finish_good where RIGHT(code, 3) = 'B05'

~~~

- Znajdź produkty w których w nazwie występuje słowo 'Premix'

~~~ sql

select * from finish_good where name like '%Premix%'

~~~

- Ile produktów pakowanych jest w określony sposób?

~~~ sql

select 

	right(code, 3) as packing_group,

	count(*) as packing_count

from finish_good

group by right(code, 3)

~~~

- Ile jest poszczególnych produktów (na podst. 2 pierwszych znaków w kodzie)

~~~ sql

select 

	left(code, 2),

	count(*)

from finish_good

group by left(code, 2)

~~~

- ile jest aktywnych i nieaktywnych klientów w poszczególnych sklepach?

~~~ sql

select 

	address.address,

	customer.active,

	count(*) as customer_count

from customer

	inner join store

		on customer.store_id = store.store_id

	inner join address

		on store.address_id = address.address_id

group by address.address, customer.active

order by address.address, customer.active

~~~

## Case

~~~ sql

select 

	first_name, 

	last_name, 

	active,

	display_active = 

	case active

		when 1 then 'Aktywny'

		when 0 then 'Nieaktywny'

	end

from customer

~~~

- przetłumacz rating filmu na język polski

~~~ sql

select 

	title, 

	rating,

	rating_pl = 

	case rating 	

		when 'NC-17' then 'od 17 lat'

		when 'PG' then 'BO z rodzicami'

		when 'PG-13' then 'od 13 lat'

		when 'G' then 'BO'

		when 'R' then 'dla dorosłych'

	end 	

from film

~~~

- Ile jest filmów w poszczególnych kategoriach wiekowych (rating)?

~~~ sql

select 

	case rating 	

		when 'NC-17' then 'od 17 lat'

		when 'PG' then 'BO z rodzicami'

		when 'PG-13' then 'od 13 lat'

		when 'G' then 'BO'

		when 'R' then 'dla dorosłych'

	end,

	count(*) as film_count

from film

group by case rating 	

		when 'NC-17' then 'od 17 lat'

		when 'PG' then 'BO z rodzicami'

		when 'PG-13' then 'od 13 lat'

		when 'G' then 'BO'

		when 'R' then 'dla dorosłych'

	end

~~~

- Uproszczenie zapytania z zastosowaniem podzapytania

~~~ sql

select 

	rating_pl,

	count(*) as film_count

from (select 

	rating_pl = case rating 	

		when 'NC-17' then 'od 17 lat'

		when 'PG' then 'BO z rodzicami'

		when 'PG-13' then 'od 13 lat'

		when 'G' then 'BO'

		when 'R' then 'dla dorosłych'

	end,

	title

from film) as film_pl

group by rating_pl

~~~

- Uproszczenie zapytania z zastosowaniem wyrazenia CTE (Common Table Expression)

~~~ sql

with film_pl as (select 

	rating = case rating 	

		when 'NC-17' then 'od 17 lat'

		when 'PG' then 'BO z rodzicami'

		when 'PG-13' then 'od 13 lat'

		when 'G' then 'BO'

		when 'R' then 'dla dorosłych'

	end,

	title

from film)

select 

	rating,

	count(*) as film_count from film_pl

group by rating

~~~

## Grupowanie po datach

- jakie mieliśmy przychody w poszczególnych latach?

~~~ sql

select

	datepart(year, payment_date) as year,

	sum(amount) as total

from payment

group by datepart(year, payment_date)

~~~

- jakie mieliśmy przychody w poszczególnych latach i miesiącach?

~~~ sql

select

	datepart(year, payment_date) as year,

	datepart(month, payment_date) as month,

	sum(amount) as total

from payment

group by 

	datepart(year, payment_date),

	datepart(month, payment_date)

order by year, month

~~~

## Funkcje okienkowe

### Narastająco

~~~ sql

select 

	amount,

	sum(amount) over (order by payment_id) as running_total

from payment

~~~

### Partycjonowanie danych

- Dochody narastająco w podziale na miesiące

~~~ sql

select 

	amount,

	datepart(month, payment_date) as month,

	sum(amount) over (partition by datepart(month, payment_date) order by payment_id) as running_total

from payment

~~~

-- ile było wypożyczeń filmów narastająco w podziale na lata

~~~ sql

select

	year,

	sum(rental_count) over (order by year) as running_total

from 

(select 

 datepart(year, rental_date) as year,

 count(*) as rental_count

from rental	

group by datepart(year, rental_date)

) as query

~~~

- z zastosowaniem wyrazeń CTE

~~~ sql

with query as (select 

 datepart(year, rental_date) as year,

 count(*) as rental_count

from rental	

group by datepart(year, rental_date)

) 

select

	year,

	sum(rental_count) over (order by year) as running_total

from query

~~~

## Zadanie

- Czy film 'Academy Dinosaur' jest dostepny w sklepie Store 1? 

~~~ sql

~~~

- Które filmy mamy 2 kopie?

~~~ sql

select 

	film.title,

	count(*) as number_of_copies 

from film

	inner join inventory

		on film.film_id = inventory.film_id

group by film.title

having count(*) = 2

~~~

- W których kategoriach filmów mamy powyżej 50 tytułów?

~~~ sql

select 

	category.name,

	count(*) as film_count

from film

	inner join film_category 

		on film.film_id = film_category.film_id

	inner join category

		on film_category.category_id = category.category_id

group by category.name

having count(*)> 50

order by film_count

~~~

- pokaż 3 kraje z których najczęściej wypożyczane są filmy

~~~ sql

select top(3)

	country.country,

	count(rental.rental_id) as rental_count

from rental

	inner join customer

		on rental.customer_id = customer.customer_id

	inner join address 

		on customer.address_id = address.address_id

	inner join city

		on city.city_id = address.city_id

	inner join country

		on city.country_id = country.country_id

	group by country.country

order by rental_count desc

~~~

- znajdź wypożyczenia, które nie zostały opłacone

~~~ sql

select distinct

	customer.first_name,

	customer.last_name

from rental	

	left outer join payment

		on rental.rental_id = payment.rental_id

	inner join customer

		on rental.customer_id = customer.customer_id

where 

	payment.payment_id is null

~~~	

# Lista funkcji

https://docs.microsoft.com/en-us/sql/t-sql/functions

https://www.tutorialspoint.com/t_sql/t_sql_functions.htm

# Pobieranie danych do Excela za pomocą zapytania SQL

https://www.mssqltips.com/sqlservertip/4585/using-microsoft-query-in-excel-to-retreive-sql-server-data/